Evaluate the definite integral. $\int^{6\pi}_{\frac{11\pi}{2}}9\sin(x)\,dx = $
Solution: First, use the sine rule: $\begin{aligned}\int^{6\pi}_{\frac{11\pi}{2}}9\sin(x)\,dx~&=~-9\cos(x)\Bigg|^{6\pi}_{{\frac{11\pi}{2}}}\end{aligned}$ Second, plug in the limits of integration: $(-9\cdot{\cos(6\pi)})-(-9\cdot{\cos(\frac{11\pi}{2})}) = -9-0 =-9$. The answer: $\int^{6\pi}_{\frac{11\pi}{2}}9\sin(x)\,dx~=~-9$